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group By hAving mAx

直接select max(num) 再group by 就行 不用后面的having

UPDATE W SET W.A = ( SELECT W.A - S.A from S where W.G = S.G and S.N in(SELECT max(N) FROM S group by G ) ) where exists (select 1 from S where W.G = S.G and S.N in(SELECT max(N) FROM S group by G ) ) exists 语句很重要,否则如...

select cno from score group by cno having degree

select a.* from table a, (select 姓名,max(开始日期) 日期 from table group by 姓名) b where a.姓名=b.姓名 and a.开始日期=b.日期

MIN(), MAX()是聚合函数. group by 后面是要跟着的 select 中所有不是聚合函数的字段。 ex1: select count(*) from emp; //只是查询总总数 emp这张表里一共有多少条记录 所以不用group by ex2: select count(*) , deptno from emp group by dept...

(1)一般而言在group by语句中,的结果显示只能是分组字段,如果有其他字段那么必须是带有聚合函数的,比如count,sum,max,min等,但是这里直接放个*就可以执行了? (2)如果是在group by之前筛选,那么怎么count,是选择oid=5的进行count么?...

1、聚合函数可以嵌套2次,所以max(avg(sal))没有问题。 2、提示ORA-00937: not a single-group group function,是因为这已经是两层关系了。进行拆分就是: select max(a.str_avg),a.deptno from (select avg(sal) as str_avg,deptno from emp g...

1 select * from (select nodecol,count(*) counts from tmp3 group by nodecol) twhere counts=(select max(count(*)) from tmp3 group by nodecol) 2 select * from tmp3 where nodecol in(select t.nodecol from (select nodecol,count(*) co...

譬如如下数据 id value 1 2 1 3 2 3 3 5 3 6 可以写个语句统计value的分组 select id,sum(value) from table group by id having sum(value)>=5 这样的结果就是 1 5 3 11 其实这句的意思就是 select id,sum(value) from table where sum(value)>...

用查询嵌套,举例: select max(a) from ( select sum(a) as a from table group by XXX ) t group by YYY

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